Let $X$ and $Y$ be random variables taking values in a separable metric space $(S,d)$. The metric I have in mind is $$\rho(X,Y) = \mathbb{E}[\min\{d(X,Y),1\}]$$ if $X$ and $Y$ take values in the a metric space $(S,d)$.

Question: Does this metric have an official name?

Wikipedia calls it the "Lévy metric for $L^0$". Also, someone once told me they thought it was called the Lévy metric. I can't find it anywhere else. (If you Google™ it, don't use anything written by me as evidence that it is called the Lévy metric!) Moveover, Lévy metric is the name for another metric.

I also know this metric metricizes convergence in probability (measure), and is equivalent to the Ky Fan metric (which I have also seen called the probability distance): $$K(X,Y) = \inf\{\varepsilon \geq 0: P(d(X,Y) > \varepsilon) \leq \varepsilon\}.$$

I have also seen this very similar metric to mine $$K^*(X,Y) = \mathbb{E}\left[\frac{d(X,Y)}{1 + d(X,Y)} \right]$$ called one of the "Ky Fan metrics".

I know I could just switch to the Ky Fan metric, but I wrote a long paper using this metric $\rho$ (calling it the Lévy metric), and I don't want to have to go through the whole thing and switch to the Ky Fan metric (and change my calculations) just because I know don't know what to properly call it. (Also, I like this metric since it is reminiscent of the $L^1$-norm.)

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    $\begingroup$ I was wondering: the random variable $(X,Y)$ is measurable with respect to the product of the Borel $\sigma$-algebras ${\cal B}(S)\otimes{\cal B}(S)$. The map $d:S\times S\to\mathtt{R}^{+}$ being continuous is measurable with respect to the Borel $\sigma$-algebra ${\cal B}(S\times S)$. Don't you need to assume something like $S$ is separable to ensure that ${\cal B}(S\times S)={\cal B}(S)\otimes{\cal B}(S)$ so that $d(X,Y)$ is a measurable map and your expectation is well-defined? $\endgroup$ Aug 11 '13 at 5:59
  • $\begingroup$ @NoelVaillant, yes indeed. I always work with separable metric spaces, so I sometimes forget to mention that assumption. I added it. $\endgroup$
    – Jason Rute
    Aug 11 '13 at 14:28

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